(a) We take x0= 0, and solve x= v0t+ 12at2(Eq. 2-15) for the acceleration: a= 2(x– v0t)/t2. Substituting x= 24.0 m, v0= 56.0 km/h = 15.55 m/s and t= 2.00 s, we find ()()222 24.0m15.55m/s2.00s3.56m/s ,2.00sa−==−or2| | 3.56 m/sa=. The negative sign indicates that the acceleration is opposite to the direction of motion of the car. The car is slowing down.
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